Module IV - Complex Variables

Definition

A function \( f(z) = u(x,y) + iv(x,y) \) is said to be analytic (or holomorphic) at a point \( z_0 \) if it is differentiable at \( z_0 \) and in some neighborhood of \( z_0 \).

2. Cauchy-Riemann Equations (Cartesian Form)

Necessary Conditions for \( f(z) \) to be Analytic: \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \]

Show that \( f(z) = z^2 \) is analytic

\( f(z) = z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy \)
So \( u = x^2 - y^2 \) and \( v = 2xy \)
Find partial derivatives:
\( u_x = 2x \), \( u_y = -2y \)
\( v_x = 2y \), \( v_y = 2x \)
Check C-R equations:
\( u_x = 2x = v_y \) ✓
\( u_y = -2y = -v_x \) ✓
Partial derivatives are continuous everywhere

C-R equations satisfied ⟹ \( f(z) = z^2 \) is analytic everywhere

Show that \( f(z) = \bar{z} = x - iy \) is not analytic

\( f(z) = x - iy \), so \( u = x \) and \( v = -y \)
Find partial derivatives:
\( u_x = 1 \), \( u_y = 0 \)
\( v_x = 0 \), \( v_y = -1 \)
Check C-R equations:
\( u_x = 1 \neq -1 = v_y \) ✗

C-R equations NOT satisfied ⟹ \( f(z) = \bar{z} \) is nowhere analytic

3. Milne-Thomson Method

Case 1: When \( u \) (real part) is given

\[ f(z) = \int \left( u_x - iu_y \right) dz + C \]

Replace \( x \to z \) and \( y \to 0 \) after finding \( u_x \) and \( u_y \)

Find analytic function \( f(z) \) if \( u = x^2 - y^2 \)

Find partial derivatives:
\( u_x = 2x \), \( u_y = -2y \)
Apply Milne-Thomson: Put \( x = z, y = 0 \):
\( u_x = 2z \), \( u_y = 0 \)
\( f(z) = \int (u_x - iu_y) \, dz = \int (2z - 0) \, dz \)
\( = z^2 + C \)

\( f(z) = z^2 + C \)

Find analytic function \( f(z) \) if \( u = e^x \cos y \)

Find partial derivatives:
\( u_x = e^x \cos y \), \( u_y = -e^x \sin y \)
Apply Milne-Thomson: Put \( x = z, y = 0 \):
\( u_x = e^z \cos 0 = e^z \)
\( u_y = -e^z \sin 0 = 0 \)
\( f(z) = \int (e^z - i \cdot 0) \, dz = \int e^z \, dz \)

\( f(z) = e^z + C \)

Case 2: When \( v \) (imaginary part) is given

\[ f(z) = \int \left( v_y + iv_x \right) dz + C \]

Replace \( x \to z \) and \( y \to 0 \) after finding \( v_x \) and \( v_y \)

Find analytic function \( f(z) \) if \( v = 2xy \)

Find partial derivatives:
\( v_x = 2y \), \( v_y = 2x \)
Apply Milne-Thomson: Put \( x = z, y = 0 \):
\( v_y = 2z \), \( v_x = 0 \)
\( f(z) = \int (v_y + iv_x) \, dz = \int (2z + 0) \, dz \)

\( f(z) = z^2 + C \)

Find analytic function \( f(z) \) if \( v = e^x \sin y \)

Find partial derivatives:
\( v_x = e^x \sin y \), \( v_y = e^x \cos y \)
Apply Milne-Thomson: Put \( x = z, y = 0 \):
\( v_y = e^z \cos 0 = e^z \)
\( v_x = e^z \sin 0 = 0 \)
\( f(z) = \int (e^z + i \cdot 0) \, dz \)

\( f(z) = e^z + C \)

4. Harmonic Function

\[ \nabla^2 \phi = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = 0 \]

i.e., \( \phi_{xx} + \phi_{yy} = 0 \)

Show that \( u = x^2 - y^2 \) is harmonic

\( u_x = 2x \), \( u_{xx} = 2 \)
\( u_y = -2y \), \( u_{yy} = -2 \)
\( u_{xx} + u_{yy} = 2 + (-2) = 0 \)

Laplace equation satisfied ⟹ \( u = x^2 - y^2 \) is harmonic

Show that \( \phi = e^x \cos y \) is harmonic

\( \phi_x = e^x \cos y \), \( \phi_{xx} = e^x \cos y \)
\( \phi_y = -e^x \sin y \), \( \phi_{yy} = -e^x \cos y \)
\( \phi_{xx} + \phi_{yy} = e^x \cos y - e^x \cos y = 0 \)

Laplace equation satisfied ⟹ \( \phi = e^x \cos y \) is harmonic

5. Harmonic Conjugate

Definition

If \( u \) and \( v \) are harmonic functions such that \( f(z) = u + iv \) is analytic, then \( v \) is called the harmonic conjugate of \( u \).

To find harmonic conjugate \( v \) given \( u \): \[ v = \int u_x \, dy - \int \left( u_y + \frac{\partial}{\partial x}\int u_x \, dy \right) dx \]

OR use Milne-Thomson method to find \( f(z) \), then separate \( v \)

Find harmonic conjugate of \( u = x^2 - y^2 \)

Using Milne-Thomson (easier):
\( u_x = 2x \), \( u_y = -2y \)
Put \( x = z, y = 0 \): \( u_x = 2z \), \( u_y = 0 \)
\( f(z) = \int (2z - 0) \, dz = z^2 + C \)
Let \( C = 0 \): \( f(z) = z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy \)
So \( v = 2xy \)

Harmonic conjugate \( v = 2xy \)

Find harmonic conjugate of \( u = e^x \cos y \)

Using Milne-Thomson:
\( u_x = e^x \cos y \), \( u_y = -e^x \sin y \)
Put \( x = z, y = 0 \): \( u_x = e^z \), \( u_y = 0 \)
\( f(z) = \int e^z \, dz = e^z + C \)
Let \( C = 0 \): \( f(z) = e^z = e^{x+iy} = e^x(\cos y + i\sin y) \)
So \( v = e^x \sin y \)

Harmonic conjugate \( v = e^x \sin y \)

6. Orthogonal Trajectories

Key Property

If \( f(z) = u + iv \) is analytic, then:

Family \( u(x,y) = c_1 \) and Family \( v(x,y) = c_2 \) are orthogonal trajectories

Method to find Orthogonal Trajectories:

Given curve: \( u(x,y) = c \)
Find \( v \) (harmonic conjugate of \( u \)) using Milne-Thomson
Orthogonal trajectories: \( v(x,y) = k \)

Find orthogonal trajectories of \( x^2 - y^2 = c \)

Given \( u = x^2 - y^2 \)
Using Milne-Thomson:
\( u_x = 2x \), \( u_y = -2y \)
Put \( x = z, y = 0 \): \( u_x = 2z \), \( u_y = 0 \)
\( f(z) = \int 2z \, dz = z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy \)
So \( v = 2xy \)

Orthogonal trajectories: \( 2xy = k \) (rectangular hyperbolas)

Find orthogonal trajectories of \( e^x \cos y = c \)

Given \( u = e^x \cos y \)
From previous example, we found \( f(z) = e^z = e^x(\cos y + i\sin y) \)
So \( v = e^x \sin y \)

Orthogonal trajectories: \( e^x \sin y = k \)

Find orthogonal trajectories of \( x^2 + y^2 = 2ax \)

Rewrite: \( x^2 - 2ax + y^2 = 0 \) → circles passing through origin
Let \( u = \frac{x^2 + y^2}{x} = x + \frac{y^2}{x} \) (but this is messy)
Better approach: Given family is \( x^2 + y^2 - 2ax = 0 \), or \( \frac{x^2+y^2}{2x} = a \)
For \( u = \frac{x}{x^2+y^2} \), we can show \( v = \frac{-y}{x^2+y^2} \)
This corresponds to \( f(z) = \frac{1}{z} = \frac{1}{x+iy} = \frac{x-iy}{x^2+y^2} \)
\( v = c \) gives \( \frac{y}{x^2+y^2} = c \) → \( x^2 + y^2 = \frac{y}{c} \) → \( x^2 + y^2 - 2by = 0 \)

Orthogonal trajectories: \( x^2 + y^2 = 2by \) (circles through origin, perpendicular family)

Concept	Formula/Condition
Cauchy-Riemann Equations	\( u_x = v_y \), \( u_y = -v_x \)
Milne-Thomson (given u)	\( f(z) = \int(u_x - iu_y)dz \) [put x=z, y=0]
Milne-Thomson (given v)	\( f(z) = \int(v_y + iv_x)dz \) [put x=z, y=0]
Laplace Equation	\( \phi_{xx} + \phi_{yy} = 0 \)
Harmonic Function	Satisfies Laplace equation
Harmonic Conjugate	\( v \) such that \( f(z) = u + iv \) is analytic
Orthogonal Trajectories	\( u = c_1 \perp v = c_2 \)

Module IV: Complex Variables

1. Analytic Function